A man slides down a telegraphic pole with an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the mass of the man be $m$. The weight of the man is $w = mg$.The forces acting on the man are his weight $mg$ acting downwards and the fricti

Vedclass · Accepted Answer

$\frac{3w}{4}$

Vedclass · Answer

(C) Let the mass of the man be $m$. The weight of the man is $w = mg$.The forces acting on the man are his weight $mg$ acting downwards and the frictional force $F$ acting upwards.According to Newton's second law of motion,the net force acting on the man is equal to the product of his mass and acceleration.Since the man is sliding down with an acceleration $a = g/4$,we have:$mg - F = ma$Substituting $a = g/4$:$mg - F = m(g/4)$$mg - F = mg/4$$F = mg - mg/4$$F = \frac{3mg}{4}$Since $w = mg$,the frictional force is $F = \frac{3w}{4}$.

$A$ man slides down a telegraphic pole with an acceleration equal to one-fourth of the acceleration due to gravity. The frictional force between the man and the pole is equal to which of the following in terms of the man's weight $w$?

Similar Questions

If $p$ and $q$ are the perpendicular distances from the origin to the straight lines $x \sec \theta - y \operatorname{cosec} \theta = a$ and $x \cos \theta + y \sin \theta = a \cos 2 \theta$,then

Which of the following statements is incorrect?

In which plant was the importance of day length for flowering first observed?

When $22.4 \ L$ of $H_{2(g)}$ is mixed with $11.2 \ L$ of $Cl_{2(g)}$,each at $S.T.P.$,the moles of $HCl_{(g)}$ formed is equal to

$A$ person who tosses an unbiased coin gains two points for turning up a head and loses one point for a tail. If three coins are tossed and the total score $X$ is observed,then the range of $X$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module